Answer :
Sure, let's start with part (a) and then move on to part (b).
### Part (a): Expressing the Half-Life \( \lambda \) in Terms of \( k \)
The half-life \( \lambda \) is defined as the time it takes for the amount of the substance to decrease to half of its initial amount \( y_0 \). The decay of the substance follows the equation:
[tex]\[ y = y_0 e^{-kt} \][/tex]
1. By definition, when \( t = \lambda \), the amount \( y \) is half of \( y_0 \), so:
[tex]\[ y = \frac{y_0}{2} \][/tex]
2. Substituting \( y \) and \( t = \lambda \) into the decay equation:
[tex]\[ \frac{y_0}{2} = y_0 e^{-k \lambda} \][/tex]
3. Divide both sides by \( y_0 \) to simplify:
[tex]\[ \frac{1}{2} = e^{-k \lambda} \][/tex]
4. Take the natural logarithm on both sides to solve for \( \lambda \):
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln\left(e^{-k \lambda}\right) \][/tex]
[tex]\[ \ln\left(\frac{1}{2}\right) = -k \lambda \][/tex]
5. Simplify \( \ln\left(\frac{1}{2}\right) \):
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2) \][/tex]
6. Substitute and solve for \( \lambda \):
[tex]\[ -\ln(2) = -k \lambda \][/tex]
[tex]\[ \lambda = \frac{\ln(2)}{k} \][/tex]
Thus, the half-life \( \lambda \) in terms of \( k \) is:
[tex]\[ \lambda = \frac{\ln(2)}{k} \][/tex]
### Part (b): Verifying the Amount at Time \( t_1 + \lambda \)
Now, we need to verify that the amount at time \( t_1 + \lambda \) is half of the amount at time \( t_1 \).
1. Suppose at time \( t_1 \), the amount is \( y_1 \). Then:
[tex]\[ y_1 = y_0 e^{-k t_1} \][/tex]
2. We want to find the amount \( y \) at time \( t = t_1 + \lambda \):
[tex]\[ y = y_0 e^{-k (t_1 + \lambda)} \][/tex]
3. Substitute \( \lambda = \frac{\ln(2)}{k} \):
[tex]\[ y = y_0 e^{-k \left(t_1 + \frac{\ln(2)}{k}\right)} \][/tex]
4. Simplify the exponent:
[tex]\[ y = y_0 e^{-k t_1} e^{-k \frac{\ln(2)}{k}} \][/tex]
[tex]\[ y = y_0 e^{-k t_1} e^{-\ln(2)} \][/tex]
5. Since \( e^{-\ln(2)} = \frac{1}{e^{\ln(2)}} = \frac{1}{2} \):
[tex]\[ y = y_0 e^{-k t_1} \cdot \frac{1}{2} \][/tex]
[tex]\[ y = \frac{1}{2} y_0 e^{-k t_1} \][/tex]
6. Recall that \( y_0 e^{-k t_1} = y_1 \):
[tex]\[ y = \frac{1}{2} y_1 \][/tex]
Therefore, at time \( t_1 + \lambda \), the amount will be half of the amount at time \( t_1 \):
[tex]\[ y = \frac{y_1}{2} \][/tex]
This confirms that the amount at time [tex]\( t_1 + \lambda \)[/tex] is indeed [tex]\( \frac{y_1}{2} \)[/tex], no matter what [tex]\( t_1 \)[/tex] is.
### Part (a): Expressing the Half-Life \( \lambda \) in Terms of \( k \)
The half-life \( \lambda \) is defined as the time it takes for the amount of the substance to decrease to half of its initial amount \( y_0 \). The decay of the substance follows the equation:
[tex]\[ y = y_0 e^{-kt} \][/tex]
1. By definition, when \( t = \lambda \), the amount \( y \) is half of \( y_0 \), so:
[tex]\[ y = \frac{y_0}{2} \][/tex]
2. Substituting \( y \) and \( t = \lambda \) into the decay equation:
[tex]\[ \frac{y_0}{2} = y_0 e^{-k \lambda} \][/tex]
3. Divide both sides by \( y_0 \) to simplify:
[tex]\[ \frac{1}{2} = e^{-k \lambda} \][/tex]
4. Take the natural logarithm on both sides to solve for \( \lambda \):
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln\left(e^{-k \lambda}\right) \][/tex]
[tex]\[ \ln\left(\frac{1}{2}\right) = -k \lambda \][/tex]
5. Simplify \( \ln\left(\frac{1}{2}\right) \):
[tex]\[ \ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2) \][/tex]
6. Substitute and solve for \( \lambda \):
[tex]\[ -\ln(2) = -k \lambda \][/tex]
[tex]\[ \lambda = \frac{\ln(2)}{k} \][/tex]
Thus, the half-life \( \lambda \) in terms of \( k \) is:
[tex]\[ \lambda = \frac{\ln(2)}{k} \][/tex]
### Part (b): Verifying the Amount at Time \( t_1 + \lambda \)
Now, we need to verify that the amount at time \( t_1 + \lambda \) is half of the amount at time \( t_1 \).
1. Suppose at time \( t_1 \), the amount is \( y_1 \). Then:
[tex]\[ y_1 = y_0 e^{-k t_1} \][/tex]
2. We want to find the amount \( y \) at time \( t = t_1 + \lambda \):
[tex]\[ y = y_0 e^{-k (t_1 + \lambda)} \][/tex]
3. Substitute \( \lambda = \frac{\ln(2)}{k} \):
[tex]\[ y = y_0 e^{-k \left(t_1 + \frac{\ln(2)}{k}\right)} \][/tex]
4. Simplify the exponent:
[tex]\[ y = y_0 e^{-k t_1} e^{-k \frac{\ln(2)}{k}} \][/tex]
[tex]\[ y = y_0 e^{-k t_1} e^{-\ln(2)} \][/tex]
5. Since \( e^{-\ln(2)} = \frac{1}{e^{\ln(2)}} = \frac{1}{2} \):
[tex]\[ y = y_0 e^{-k t_1} \cdot \frac{1}{2} \][/tex]
[tex]\[ y = \frac{1}{2} y_0 e^{-k t_1} \][/tex]
6. Recall that \( y_0 e^{-k t_1} = y_1 \):
[tex]\[ y = \frac{1}{2} y_1 \][/tex]
Therefore, at time \( t_1 + \lambda \), the amount will be half of the amount at time \( t_1 \):
[tex]\[ y = \frac{y_1}{2} \][/tex]
This confirms that the amount at time [tex]\( t_1 + \lambda \)[/tex] is indeed [tex]\( \frac{y_1}{2} \)[/tex], no matter what [tex]\( t_1 \)[/tex] is.
